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Let $V$ be the vector space of real $2\times 2$ symmetric matrices $X=\begin{bmatrix}x&y\\y&z\end{bmatrix}$ and let $A=\begin{bmatrix}2&1\\0&1\end{bmatrix}$. Determine the matrix of the linear operator on $V$ defined by $X\rightsquigarrow AXA^t$ with respect to a suitable basis.

So far, I got $AXA^t=\begin{bmatrix}4x+4y+z&2y+z\\2y+z&z\end{bmatrix}$

I need to find a matrix $C$ such that $CX=AXA^t$, but not find $C$. Am I doing anything wrong. Please someone tell me what is wrong here.

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  • $\begingroup$ I think you have a small misunderstanding about the property that $C$ should satisfy. Letting $C$ be the matrix of your linear operator, then $C$ satisfies $C[X]_\mathcal{B} = [AXA^t]_\mathcal{B}$, where $[X]_\mathcal{B}$ is the coordinate vector of $X$ with respect to a basis $\mathcal{B}$ for $V$. Since $V$ is $2$-dimensional (Why? Can you find a basis for $V$?), then $[X]_\mathcal{B} \in \mathbb{R}^2$ for each $X \in V$, and $C$ is a $2 \times 2$ matrix. $\endgroup$ – André 3000 Oct 21 '17 at 13:34
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    $\begingroup$ @Quasicoherent $V$ is $3$-dimension, isn't? Basis is $\{\begin{bmatrix}1&0\\0&0\end{bmatrix}, \begin{bmatrix}0&1\\1&0\end{bmatrix}, \begin{bmatrix}0&0\\0&1\end{bmatrix} \}$. $\endgroup$ – algebra_001 Oct 21 '17 at 13:57
  • $\begingroup$ Whoops, you're right: my mistake! Sorry, I missed that $x$ and $z$ were different. The rest of my comment is still true, though, except that $[X]_\mathcal{B} \in \mathbb{R}^3$ and $C$ should be $3 \times 3$. $\endgroup$ – André 3000 Oct 21 '17 at 14:11
  • $\begingroup$ @Quasicoherent , it just hit me hard!!! Using the basis I wrote, the co-ordinate of V should be $(x,y,z)$, isn't? But $(x,y,z)$ is not even a element of $V$.... how is that possible! $\endgroup$ – algebra_001 Oct 21 '17 at 17:01
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First, some generalities. The structure of a vector space is totally determined by its dimension. Given a real vector space $V$ of dimension $n$, then $V$ is isomorphic to $\mathbb{R}^n$, where the isomorphism is given by any choice of basis for $V$. More explicitly, given a basis $\newcommand{\B}{\mathcal{B}} \newcommand{\R}{\mathbb{R}} \B = \{v_1, \ldots, v_n\}$ for $V$, we get an isomorphism given by the coordinate map \begin{align*} \varphi_\B: V &\to \R^n\\ x &\mapsto [x]_\B \end{align*} where $[x]_\B \in \R^n$ is the coordinate vector of the vector $x \in V$, whose entries are the unique weights $c_1, \ldots, c_n \in \R$ expressing $x$ as a linear combination of $v_1, \ldots, v_n$. That is, $$ [x]_\B = \begin{pmatrix} c_1\\ \vdots\\ c_n \end{pmatrix} $$ where $$ x = c_1 v_1 + \cdots + c_n v_n \, . $$ Once we've chosen a basis and identified with $\R^n$, a linear map $T: V \to V$ can be expressed as left-multiplication by some matrix $[T]_\B$. This is captured by the following commutative diagram. $\hspace{6.5cm}$

As for computing $[T]_\B$, one can show that $$ [T]_\B = \begin{pmatrix} | & & |\\ [T(v_1)]_\B & \cdots & [T(v_n)]_\B\\ | & & | \end{pmatrix} $$ where as before $\B = \{v_1, \ldots, v_n\}$.

Returning to your problem, you have found a basis for $V$, and also determined the general formula for $T$. Now you just have to apply $T$ to the basis vectors, write each image as a linear combination of the basis vectors, and use these as the columns of $[T]_\B$. For instance, your first basis vector is $v_1 = \begin{pmatrix}1 & 0\\ 0 & 0\end{pmatrix}$ and $$ T(v_1) = \begin{pmatrix}4 & 0\\ 0 & 0\end{pmatrix} = 4 v_1 $$ so $[T(v_1)]_\B = \begin{pmatrix}4\\ 0\\ 0\end{pmatrix}$. Can you compute the remaining columns?

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    $\begingroup$ Yeah! I actually got the answer. I found $$\begin{bmatrix}4&0&0\\4&2&0\\1&1&1\end{bmatrix}$$ $\endgroup$ – algebra_001 Oct 23 '17 at 4:28
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    $\begingroup$ That's almost right, but you've actually written the transpose of the matrix. You should write the coordinate vectors as the columns of the matrix, not the rows (unless your conventions differ from mine). $\endgroup$ – André 3000 Oct 23 '17 at 5:09

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